A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of 15 MJ/MVA. If the input and output powers of the alternator are 15 MW and 10 MW, respectively, the angular acceleration in mechanical degree/ s² is _______. (round off to nearest integer)

The correct answer is 75.

Step-by-step Explanation:

First, let us identify the given parameters from the problem:
Rating of the alternator (or machine base), $S=20\text{MVA}$
Inertia constant, $H=15\text{MJ/MVA}$
Number of poles, $P=4$
Frequency, $f=50\text{Hz}$
Mechanical input power, $P_m=15\text{MW}$
Electrical output power, $P_e=10\text{MW}$

The accelerating power ($P_a$) is the difference between the input mechanical power and the output electrical power:
$P_a=P_m-P_e=15\text{MW}-10\text{MW}=5\text{MW}$

According to the swing equation:
$M\frac{d^2\delta }{d{t}^2}=P_m-P_e=P_a$
where $M$ is the angular momentum (or inertia coefficient) and $\frac{d^2\delta }{d{t}^2}$ is the electrical angular acceleration (${\alpha }_e$).

The inertia coefficient $M$ in MJ-s/electrical degree is given by the formula:
$M=\frac{S\times H}{180\times f}$

Substituting the given values into the formula for $M$:
$M=\frac{20\times 15}{180\times 50}=\frac{300}{9000}=\frac{1}{30}\text{MJ-s/electrical degree}$

Now, we can find the electrical angular acceleration (${\alpha }_e$):
${\alpha }_e=\frac{P_a}{M}=\frac{5}{1/30}=150{\text{electrical degree/s}}^2$

The relation between electrical degrees and mechanical degrees is:
${\theta }_e=\frac{P}{2}{\theta }_m$
Therefore, the mechanical angular acceleration (${\alpha }_m$) is:
${\alpha }_m={\alpha }_e\times \frac{2}{P}$

Substituting ${\alpha }_e=150$ and $P=4$:
${\alpha }_m=150\times \frac{2}{4}=75{\text{mechanical degree/s}}^2$